\newproblem{lay:6_4_22}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 6.4.22}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $\{\mathbf{u}_1,...,\mathbf{u}_p\}$ be an orthogonal basis for a subspace $W$ of $\mathbb{R}^n$, and let $T:\mathbb{R}^n\rightarrow \mathbb{R}^n$ be
	defined by $T(\mathbf{x})=\mathrm{Proj}_W\{\mathbf{x}\}$. Show that $T$ is a linear transformation.
}{
   % Solution
	Since $\{\mathbf{u}_1,\mathbf{u}_2,...,\mathbf{u}_p\}$ is an orthogonal basis of $W$, the projection onto $W$ can be calculated as
	\begin{center}
		$T(\mathbf{x})=\mathrm{Proj}_W\{\mathbf{x}\}=\frac{\mathbf{x}\cdot\mathbf{u}_1}{\mathbf{u}_1\cdot\mathbf{u}_1}\mathbf{u}_1+...+
		   \frac{\mathbf{x}\cdot\mathbf{u}_p}{\mathbf{u}_p\cdot\mathbf{u}_p}\mathbf{u}_p$
	\end{center}
	To show that $T$ is linear let us show that $T(c\mathbf{x})=cT(\mathbf{x})$
	\begin{center}
		$\begin{array}{rcl}
			T(c\mathbf{x})&=&\frac{(c\mathbf{x})\cdot\mathbf{u}_1}{\mathbf{u}_1\cdot\mathbf{u}_1}\mathbf{u}_1+...+
		   \frac{(c\mathbf{x})\cdot\mathbf{u}_p}{\mathbf{u}_p\cdot\mathbf{u}_p}\mathbf{u}_p\\
			&=&c\frac{\mathbf{x}\cdot\mathbf{u}_1}{\mathbf{u}_1\cdot\mathbf{u}_1}\mathbf{u}_1+...+
		   c\frac{\mathbf{x}\cdot\mathbf{u}_p}{\mathbf{u}_p\cdot\mathbf{u}_p}\mathbf{u}_p\\
			&=&c\left(\frac{\mathbf{x}\cdot\mathbf{u}_1}{\mathbf{u}_1\cdot\mathbf{u}_1}\mathbf{u}_1+...+
		       \frac{\mathbf{x}\cdot\mathbf{u}_p}{\mathbf{u}_p\cdot\mathbf{u}_p}\mathbf{u}_p\right)\\
			&=&cT(\mathbf{x})
		\end{array}$
	\end{center}
	and that $T(\mathbf{x}_1+\mathbf{x}_2)=T(\mathbf{x}_1)+T(\mathbf{x}_2)$
	\begin{center}
		$\begin{array}{rcl}
			T(\mathbf{x}_1+\mathbf{x}_2)&=&\frac{(\mathbf{x}_1+\mathbf{x}_2)\cdot\mathbf{u}_1}{\mathbf{u}_1\cdot\mathbf{u}_1}\mathbf{u}_1+...+
		   \frac{(\mathbf{x}_1+\mathbf{x}_2)\cdot\mathbf{u}_p}{\mathbf{u}_p\cdot\mathbf{u}_p}\mathbf{u}_p\\
			&=&\frac{\mathbf{x}_1\cdot\mathbf{u}_1+\mathbf{x}_2\cdot\mathbf{u}_1}{\mathbf{u}_1\cdot\mathbf{u}_1}\mathbf{u}_1+...+
		   \frac{\mathbf{x}_1\cdot\mathbf{u}_p+\mathbf{x}_2\cdot\mathbf{u}_p}{\mathbf{u}_p\cdot\mathbf{u}_p}\mathbf{u}_p\\
		  &=&\left(\frac{\mathbf{x}_1\cdot\mathbf{u}_1}{\mathbf{u}_1\cdot\mathbf{u}_1}\mathbf{u}_1+...+
						   \frac{\mathbf{x}_1\cdot\mathbf{u}_p}{\mathbf{u}_p\cdot\mathbf{u}_p}\mathbf{u}_p\right)+
				 \left(\frac{\mathbf{x}_2\cdot\mathbf{u}_1}{\mathbf{u}_1\cdot\mathbf{u}_1}\mathbf{u}_1+...+
						   \frac{\mathbf{x}_2\cdot\mathbf{u}_p}{\mathbf{u}_p\cdot\mathbf{u}_p}\mathbf{u}_p\right)\\
			&=&T(\mathbf{x}_1)+T(\mathbf{x}_2)
		\end{array}$
	\end{center}
	
}
\useproblem{lay:6_4_22}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
